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If cot θ = 7/8, evaluate : (i) (1 + sin θ)(1 - sin θ)/(1+cos θ)(1-cos θ) (ii) cot^2 θ

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one month ago Anand Kumar Pandey
1372 Points
```							Dear StudentLet us assumea△ABC in which∠B = 90° and∠C= θGiven:cotθ= BC/AB = 7/8Let BC = 7k and AB = 8k,where k is a positive real numberAccording to Pythagoras theorem in△ABCwe getAC^2= AB^2+BC^2AC^2= (8k)^2+(7k)^2AC^2= 64k^2+49k^2AC^2= 113k^2AC =√113 kAccording to the sine and cos function ratios,it is written assinθ = AB/AC= Opposite Side/Hypotenuse= 8k/√113 k= 8/√113andcosθ=Adjacent Side/Hypotenuse= BC/AC= 7k/√113 k= 7/√113Now apply the values of sin function and cos function:i)(1 + sin θ)(1 - sin θ)/(1+cos θ)(1-cos θ)= 1-sin^2θ/1-cos^2θ   = cos^2θ/sin^2θ =49/64ii) cot^2 θ=cos^2θ/sin^2θ =49/64Thanks
```
one month ago
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